For the same operator algebra people: see also Alain Connes, A factor not anti-isomorphic to itself Ann. Math. (2) 101, 1975, 536-554.
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Alain ValetteMay 9 '11 at 11:23

1

Out of curiosity: why do operator people publish papers about these examples? Are they much rarer there? Us mere algebrist don't think much of this, I think (all my algebra undergrad students have had to check that some example or another works)
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Mariano Suárez-Alvarez♦May 9 '11 at 15:21

7

@Mariano: It is extraordinarily hard to see whether von Neumann algebras are isomorphic. Results like these are usually hard-won and are accompanied by deep new understanding of the structure of the algebras. (Notice that the above paper of Connes is an Annals paper.)
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Jon BannonMay 9 '11 at 15:34

Thanks everyone for the nice answers. I would like to have accepted more than one.
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Amritanshu PrasadMay 10 '11 at 10:36

2

The $*$-operator algebra examples are non-examples for this question: they have a involution, so are rings isomorphic to their opposite rings. The point of the (difficult) examples is that they have no complex-linear anti-automorphism as rings with involution i.e. complex-linear anti-automorphisms that commute with the fixed anti-automorphism $*$ (b.t.w. in $C^*$ algebras all ring [anti]automorphism are real-linear, but not always complex-linear)
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user46855Feb 15 '14 at 15:45

8 Answers
8

Here's a factory for making examples. If $\Gamma$ is a quiver, and $k$ a field, then we get a quiver algebra $k\Gamma$. If $\Gamma$ has no oriented cycles, we can recover $\Gamma$ from $k\Gamma$ by taking the Ext-construction. Also, the opposite algebra of a quiver algebra is obtained by reversing all the arrows in the quiver.

Hence you can produce an example by taking the quiver algebra of any quiver with no oriented cycles, which is not isomorphic to its reverse. It's easy to construct lots of quivers with these properties.

It's clear if you believe that theorem, that you can recover the original quiver!
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James CranchMay 9 '11 at 14:16

2

Incidentally, the smallest such quiver is the V-shaped quiver with two arrows emanating from one point. This gives a five-dimensional algebra: that's quite neat and tidy!
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James CranchMay 9 '11 at 15:09

2

First, one finds the vertices of the quiver $\Gamma$, by finding the unique maximal set of orthogonal idempotents $\{e_i\}$; these will correspond to vertices. Then, the number of arrows from $i$ to $j$ will be $Ext^1(\Gamma e_j,\Gamma e_i)$.
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Greg MullerMay 10 '11 at 15:49

1

Alternatively: since the algebra is finite dimensiona, there are a finite number of isomorphism classes of simple modules. Use them as vertices of a graph, and draw $\dim\operatorname{Ext}^1(S,T)$ arrows from the class of $S$ to the class of $T$.
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Mariano Suárez-Alvarez♦May 10 '11 at 16:00

1

Whoops, sorry, I confused two different ways of getting the arrows. The $Ext^1$ should be between $k e_j$ and $ke_i$, the simple modules supported at the appropriate vertices. You can also get the arrows as degree 1 graded $Hom$ between projectives $\Gamma e_i$ and $\Gamma e_j$, but the $Ext$ between them certainly vanishes.
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Greg MullerMay 10 '11 at 16:05

Here is an easy example. Consider the abelian group $M = \mathbb{Z} \times \mathbb{Q}$. I claim that $R:=\text{End}(M)$ does not have any anti-endomorphism at all. EDIT: My previous proof is flawed. Thanks to Leon Lampret who pointed this out to me. The new proof shows that $R$ has several anti-endomorphisms, but no one is invertible. Thus $R$ is not isomorphic to $R^{\mathrm{op}}$.

To amplify on Bugs Bunny's answer: let $D$ be a finite dimensional central division algebra over a field $K$. Then $D \otimes_K D^{\operatorname{op}} \cong \operatorname{End}_K(D)$. From this it follows that in the Brauer group of $K$, the class of $D^{\operatorname{op}}$ is the inverse of the class of $D$. So a central division algebra over a field is isomorphic to its opposite algebra iff it has order $2$ in the Brauer group, or, in the lingo of that field, period $2$.

So you can get examples by taking any field $K$ with $\operatorname{Br}(K) \neq \operatorname{Br}(K)[2]$. In particular the Brauer group of any non-Archimedean locally compact field is $\mathbb{Q}/\mathbb{Z}$ and the Brauer group of any global field is close to being the direct sum of the Brauer groups of its completions (there is one relation, the so-called reciprocity law, which says that a certain "sum of invariants" map is zero). So for instance a division algebra of dimension $9$ over its center will do and these things can be constructed over the above fields.

If you don't want to restrict to such nice algebras, it becomes much easier to construct examples. (Thanks to L. Moret-Bailly and Todd Trimble for fixing this!) Let $S$ be a set with at least two elements, and consider
the left semigroup on $S$: for all $x,y \in S$, $x \cdot y = x$. For any field $K$, let $K[S]$ be
the semigroup algebra, and let $A = K[S]^1 = K \oplus K[S]$ be the associated unital algebra. Then, if I am not still mistaken, it should be clear that $A$ is not isomorphic to its opposite algebra.

(Since I know nothing about operator algebras, my opinion is of course going to be that such examples are much more complicated than the ones given above...)

Unless I miss something, this $L(V)$ is not a ring: we have $x.(y+z)=x$ while $(x.y)+(x.z)=x+x$.
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Laurent Moret-BaillyMay 9 '11 at 12:17

True, but this can be repaired by passing to a monoid algebra for a monoid $M$ with a similar definition for multiplication (and making due allowance for the identity).
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Todd Trimble♦May 9 '11 at 12:32

Whoops! Good grief, thanks for catching this. I tried to repair it along the lines of Todd's comment, but let me know if it still seems fishy. (I'm off to delete something from my notes...)
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Pete L. ClarkMay 9 '11 at 12:43

Your example is not simple, i.e., it is not a simple algebra! If you want a simple algebra, you need a field whose Brauer group has elements of order more than 2 (the opposite algebra = inversion in Brauer group). If I remember correctly, the p-adic field will do the trick...

A general idea to construct rings which behave different on the left and on the right is the following, which is already contained in Martins's answer:
One considers triangular rings
$$ A=\begin{pmatrix} R & M \\
0 & S
\end{pmatrix}
$$
where $R$ and $S$ are rings and $M$ is an $R$-$S$-bimodule. The left and right ideals of such a ring can be decribed: for example, the left ideals are isomorphic to $U\oplus J$, where $J $ is a left ideal of $S$, and $U$ an $R$-submodule of $R\oplus M$ with $MJ \subseteq U$. (See Lam's book A First Course in Noncommutative Rings, §1) Suitable choices of $R$, $M$ and $S$ lead to examples with quite different left and right structure.
For example, the finite ring
$$\begin{pmatrix}
\mathbb{Z}/4\mathbb{Z} & \mathbb{Z}/2\mathbb{Z} \\
0 & \mathbb{Z}/2\mathbb{Z}
\end{pmatrix}$$
has 11 left ideals and 12 right ideals, if my counting is right. (This may be the smallest example of a unital ring not isomorphic to its opposite ring, but I'm not sure here.)

Of course, there are lots of examples, since there are many ring theoretic notions which are known to be not left-right symmetric. T. Y. Lam, in his two books (First Course mentioned above and Lectures on Modules and Rings), usually contructs at least one example of a ring being left blah but not right blah, whenever blah is a property which is not left-right symmetric. (Lam's books are generally worth reading, in particular when looking for examples!)

I don't know whether or not there are others of order 16 (well, aside from its opposite), but there are definitely none of order smaller than 16.
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Harry AltmanMay 9 '11 at 22:03

Thanks for pointing out Lam's book. I looked at it and indeed it is worth reading. Jacobson's example looks suspiciously similar to a triangular ring construction.
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Amritanshu PrasadMay 10 '11 at 6:15

This is a bit late, but it's my favorite class of examples. If $X$ is a smooth affine variety over $\mathbb{C}$ (say), and $\mathcal{D} = \mathcal{D}(X)$ is its algebra of differential operators, then the opposite algebra $\mathcal{D}^{op}$ is isomorphic to
$\mathcal{D}(K) = K\otimes \mathcal{D}\otimes K^{-1}$, where $K$ denotes the canonical module of $X$. [This is also true when $X$ is Gorenstein but not necessarily smooth---see work of Yekutieli.]

So one gets answers to your question when $X$ doesn't have trivial canonical bundle. [And of course the story sheafifies for any smooth variety.]

EDIT: I was writing carelessly the first time (thanks to Amri's comment for highlighting this). Note that $\mathcal{D}(K)$ acts on $K$ on the left. Since a left $\mathcal{D}$-module structure on a vector bundle (finitely generated projective module) is the same as a flat connection, one has $\mathcal{D}\cong \mathcal{D}(K)$ if and only if $K$ admits a flat connection. The first chern class of $K$ is an obstruction to the existence of a flat connection. So just pick your favorite such affine variety (see also
this MO question for discussion of that).
A pretty complete discussion of the (non)triviality of rings of twisted differential operators (TDOs) can be found in Beilinson-Bernstein "A proof of Jantzen conjectures."

This story also illuminates a little bit why differential operators on half-densities, i.e.
$\mathcal{D}(K^{1/2}) = K^{1/2}\otimes \mathcal{D}\otimes K^{-1/2}$, plays a special role in the study of rings of differential operators and (twisted) $\mathcal{D}$-modules (it's canonically isomorphic to its opposite algebra).

Here is an explicit example of a central simple algebra over $\mathbb{Q}$ not isomorphic to its opposite (which is merely a detailed example of what Pete explained).

First take a cubic cyclic Galois extension $L/\mathbb{Q}$, for instance $L = \mathbb{Q}[x] / (x^3 + x^2 − 2x − 1)$, and let $\rho$ be a non-trivial element of $\operatorname{Gal}(L/\mathbb{Q})$. Now take an arbitrary element $\gamma \in \mathbb{Q}$ which is not the norm of an element in $L$. Define
$$D = L \oplus zL \oplus z^2L,$$
where $z$ is a new "symbol" subject to the relations $z^3 = \gamma$ and $zt = t^\rho z$ for all $t \in L$.
Then $D$ is a central simple division algebra of degree $3$ (i.e. of dimension $9$), and since its image in $\operatorname{Br}(\mathbb{Q})$ has order $3$, it is not isomorphic to its opposite.

As you can imagine, this procedure works for any field admitting a cyclic extension (of degree $>2$) for which the norm is non-surjective.